jquery ajaxSubmit 异步提交的简单实现

Posted 2023-05-05

参考技术A 这篇文章主要介绍了jquery
ajaxSubmit
异步提交的简单实现。需要的朋友可以过来参考下，希望对大家有所帮助
前台js

代码如下:
$("#nickForm").ajaxSubmit(

type:
"post",

url:
"http://localhost:8080/test/myspace.do?method=updateNick¶m=1",

dataType:
"json",

success:
function(result)

//返回提示信息

alert(result.nickMsg);


);

后台封装：
代码如下:
public
ActionForward
toUpdateNickName(ActionMapping
mapping,
ActionForm
form,

HttpServletRequest
request,
HttpServletResponse
response)

PrintWriter
pw
=
response.getWriter();

JSONObject
obj
=
new
JSONObject();

obj.put("nickMsg",
"昵称修改成功！");

pw.print(obj);

pw.close();

jQuery中使用ajaxSubmit提交表单

1.http://www.cnblogs.com/liuhongfeng/p/5150389.html

2.http://blog.csdn.net/baidu_29119747/article/details/51395736